An iron nail rusts when exposed to oxygen.

Question

 How many grams of iron are needed to completely consume 14.5 L of oxygen gas according to the following reaction at 25 °C and 1 atm?

iron ( s ) + oxygen ( g )  iron(III) oxide ( s )

??grams iron

Answer 1

4Fe(s) + 3O₂(g) ==> 2Fe₂O₃(s) ... balanced equation for iron + oxygen ->iron(III) oxide

Since the experiment is conducted at 25ºC = 298K and 1 atm pressure, we use these values and the ideal gas law to find the moles of O₂ present

PV = nRT

n = PV/RT = (1 atm)(14.5 L)/(0.08214 Latm/Kmol)(298K)

n = 0.592 moles O₂

Next, we use stoichiometry of the balanced equation to find the moles and mass of Fe needed....

0.592 moles O₂ x 4 mol Fe/3 mol O₂ x 55.8 g/mol Fe = 44.1 g Fe needed (to 3 significant figures)